C Physics question I Chegg. X Physics Archive October x Wiley PLUS C edugen.wile

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C Physics question I Chegg. X Physics Archive October x Wiley PLUS C edugen.wileyplus.com/edugen/student/mainfr.uni Gradebook ORION Home Read, Study & Practice Assignment Assignment Open Assignment FULL SCREEN PRINTER VERSION BACK ONEXTD ASSIGNMENT RESOURCES Chapter 08, Problem 021 Homework 8 Fall The figure shows a pendulum of length L 2.3 m. Its bob (which effectively has all the mass) has speed vo when the cord makes an angle 34o with the vertical. (a) What is the 2015 Chapter 08. Problem speed of the bob when it is in its lowest position if vo 5.6 m/s? What is the least value that vo can have if the pendulum is to swing down and then up (b) to a horizontal position 001 and (c) to a vertical position with the cord remaining straight? d) Do the answers to (b) and (c) increase, decrease, or remain the same if eo is increased by a few degrees? Chapter 08 Problem 002 Chapter 08 Problem 006 Chapter 08 Problem 007 Chapter 08, Problem 016 Chapter 08 Problem 08, Prob 021 Chapter 08. Problem 027 Chapter 08 Problem 030 Chapter 08 Problem 031 Chapter 08 Problem (a) Number Units 036 e Chapter 08, Problem b) Number Units 038 08, Prob (c) Number Units 039 08, Prob (d) 057 Chapter 08, Problem License Agreement l Privacy Policy I G 2000-2015 John Wiley & Sons Inc. All Rights Reserved. A Division of John Wiley Sons Inc. Version 4.16.1.7 3:28 PM Ask me anything 0/18/2015

Explanation / Answer

At Inital Position,
K.Ein = 0.5*m*5.6^2
P.Ein = m*g*h , where h = L(1 - cos(34))

At Final Positon,
K.Efi = 0.5*m*v^2
P.Efi = 0

Using Energy Conservation,
K.E in + P.E in = K.E fi + P.Efin
0.5*m*5.6^2 + m*g* L(1 - cos(34)) = 0.5*m*v^2 + 0
15.68 + 9.8*2.3(1 - cos(34)) = 0.5*v^2
v = 6.25 m/s
Speed of the bob at lowest point, v = 6.25 m/s

b)
At Horizontal positon, h = 2.3 m
Using Energy Conservation,
K.E in + P.E in = K.E fi + P.Efin
0.5*m*(Vo)^2 + m*g* L(1 - cos(34)) = 0 + m*g*2.3
0.5*(Vo)^2  + 9.8*2.3(1 - cos(34)) = 2.3*9.8
Vo = 6.11 m/s
Least Value, Vo = 6.11 m/s

c)
At Vertical positon, h = 4.6 m
Using Energy Conservation,
K.E in + P.E in = K.E fi + P.Efin
0.5*m*(Vo)^2 + m*g* L(1 - cos(34)) = 0 + m*g*4.6
0.5*(Vo)^2  + 9.8*2.3(1 - cos(34)) = 4.6*9.8
Vo = 9.1 m/s
Least Value, Vo = 9.1 m/s

d)
If angle is increased by few degrees,
Least Value of Vo in (b) & (c) would Decrease.

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